http://acm.hdu.edu.cn/showproblem.php?pid=4786 Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 49    Accepted Submission(s): 26 Problem Description Coach Pang is interested in Fi…

HDU 3117 Fibonacci Numbers(斐波那契前后四位,打表+取对+矩阵高速幂) ACM 题目地址:HDU 3117 Fibonacci Numbers 题意:  求第n个斐波那契数的前四位和后四位.  不足8位直接输出. 分析:  前四位有另外一题HDU 1568,用取对的方法来做的.  后四位能够用矩阵高速幂,MOD设成10000即可了. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.c…

一只小蜜蜂... Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description 有一只经过训练的蜜蜂只能爬向右侧相邻的蜂房,不能反向爬行.请编程计算蜜蜂从蜂房a爬到蜂房b的可能路线数. 其中,蜂房的结构如下所示. Input 输入数据的第一行是一个整数N,表示测试实例的个数,然后是N 行数据,每行包含两个…

首先计算图的联通情况,如果图本身不联通一定不会出现生成树,输出"NO",之后清空,加白边,看最多能加多少条,清空,加黑边,看能加多少条,即可得白边的最大值与最小值,之后判断Fibonacci数是否在这两个之间,如果是输出yes,否则no. 然而,,然而,,我看的题解有问题!!!!!调了俩小时愣是没找出错误来,,然后把题解交了发现过不了,,,,真是够了.,.,., 第二天上午终于A了,,满分程序: #include <cstdio> #include <cstring&…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 75    Accepted Submission(s): 38 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do…

http://acm.hdu.edu.cn/showproblem.php? pid=4786 Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1733    Accepted Submission(s): 543 Problem Description Coach Pang is interested i…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2952    Accepted Submission(s): 947 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him t…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4786 题意:有N个节点(1 <= N <= 10^5),M条边(0 <= M <= 10^5).当中一部分边被染成了黑色,剩下的边是白色,问能不能建立一棵树,树中有斐波那契数个白色边. 思路:用克鲁斯卡尔建三次树,第一是用全部边建树.推断能否建成一棵树,第二次用黑边建树,最多能够用到x条黑边(不成环),n-1-x就是最少须要用的白边的数量,第三次用白边建树,最多能够用到y条白边.假设在[y…

Pandigital Fibonacci ends The Fibonacci sequence is defined by the recurrence relation: F[n] = F[n-1] + F[n-2], where F[1] = 1 and F[2] = 1. It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digi…

// 面试题:斐波那契数列 // 题目:写一个函数,输入n,求斐波那契(Fibonacci)数列的第n项. #include <iostream> using namespace std; // ====================方法1:递归==================== //注意这种递归方法虽然看起来很简单,但是由于压入栈和弹出,会存在栈溢出的可能,而且效率特别慢,且n越大效率越慢 long long Fibonacci_Solution1(unsigned int n)//…